Question

A line M through A is drawn parallel to BD. Point S moves such that its distances from the line BD and the vertex A are equal. If locus of S cuts M at $T_2$ and $T_3$ and AC at $T_1$, then area of $\Delta T_1{T}_2{T}_3$ is

• A. $\frac{1}{2}$ sq. units
• B. $\frac{2}{3}$ sq. units
• C. $1$ sq. unit
• D. $2$ sq. units

Answer 1

Answer: (C)

$1$ sq. unit

Since, $AO=\sqrt{2}$

$\therefore A{T}_1=T_{1}O=\frac{1}{\sqrt{2}}$

As $A$ is the focus, $T_1$ is the vertex and $BD$ is the directrix of parabola.

Also, $T_2{T}_3$ is latus rectum

$\therefore T_2{T}_3=4.\frac{1}{\sqrt{2}}$

$\therefore$ Area of $\Delta T_1{T}_2{T}_3=\frac{1}{2}\times \frac{1}{\sqrt{2}}\times \frac{4}{\sqrt{2}}=1$ sq. unit